[leetcode 224 & 227]Basic Calculator I & II 原创解法

题目概述

224原题链接

227原题链接

Implement a basic calculator to evaluate a simple expression string.

这两道题可以使用一种通用的解法来解。所以就放在一起了。

"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23

"3+2*2" = 7
" 3/2 " = 1
" 3+5 / 2 " = 5

波兰计数法

波兰计数法是一种非常巧妙的方法，只使用一个栈结构就能计算表达式。 但是呢，我们常用的表达式是一种infix表达式，而波兰计数法是一种 postfix表达式。所以需要一种把infix expression转换成postfix expression 。转换方法

抽象语法树

我还是想用抽象语法树的形式来解。也就是说先把表达式转换成一棵二叉树。 然后再计算这棵树。思路很简单，所以呢，为了直观的看到树状结构。 我发现Python下有一个不错的包binarytree，这个包可以打印树状图。 我们先来看看打印的结果吧。是不是很不错。

0-(1+2)*3*(4-3/5/5)+1+1

                                __+  
                               /   \ 
    __________________________+     1
   /                           \     
  -__________                   1    
 /           \                       
0           __*__                    
           /     \                   
        __*       -______            
       /   \     /       \           
      +     3   4       __/          
     / \               /   \         
    1   2             /     5        
                     / \             
                    3   5

初步解法

根据以上的思路，我们首先定义一个Tree，继承自binarytree的Node。 然后就先对字符串处理成token流，其实这题没这个要求。简单的去掉 空格就行了。也没有复杂的数字。下面给出了生成AST的初步代码。 算是递归下降法吧。 然后再计算树的值。

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# python2
from binarytree import Node

class TreeNode(Node):
    def __init__(self, value=None):
        super(TreeNode, self).__init__(value)

    def compute(self, root):
        if root:
            if root.left == None and root.right == None:
                return int(root.value)
            else:

                left = self.compute(root.left)
                right = self.compute(root.right)
                result = None
                char = root.value
                if char == "+":
                    result = left + right
                elif char == "-":
                    result = left - right
                elif char == "*":
                    result = left * right
                elif char == "/":
                    result = left / right

                return result

class Test(object):
    def __init__(self, s):
        self.s = self.tokenize(s + "$")
        # print(self.s)
        self.i = 0
        self.n = len(self.s)
        self.t = None
        self.precedences = {'+': 0, '-': 0, '*': 1, '/': 1}

    def tokenize(self, s):
        n = len(s)
        result = []
        tmp = ""
        for i in range(0, n):
            char = s[i]
            if char == ' ':
                continue
            if char in '+-*/()$':
                if tmp:
                    result.append(tmp)
                    tmp = ""
                result.append(char)
                continue
            if char in [str(i) for i in range(0, 10)]:
                tmp += char

        return result

    def parse(self):
        while self.s[self.i] != '$':
            self.parse_expr()
        return self.t

    def parse_expr(self):
        while True:
            char = self.s[self.i]
            t = None
            if char in '$)':
                break
            elif char in '(':
                self.i += 1
                self.t = self.parse_expr()
                c = self.s[self.i]
                if c != ")":
                    raise Exception("{0} shoud be )".format(self.i))
                self.i += 1
            elif char in "+-":
                lhs = self.t
                self.t = self.parse_binary_operator(lhs, self.precedences[char])
            elif char in "*/":
                lhs = self.t
                self.t = self.parse_binary_operator(lhs, self.precedences[char])
            else:
                self.t = self.parse_term()
        return self.t

    def parse_term(self):
        c = self.s[self.i]
        if c == "(":
            self.i += 1
            t = self.parse_expr()
            # print(t)
            c = self.s[self.i]
            if c != ")":
                raise Exception("{0} shoud be )".format(self.i))
            else:
                self.i += 1
                return t
        else:
            t = TreeNode(c)
            self.i += 1
            return t

    def parse_binary_operator(self, lhs, precedence):
        c = self.s[self.i]
        oper = TreeNode(c)
        self.i += 1
        oper.left = lhs
        rhs = self.parse_term()
        while True:
            lookahead = self.s[self.i]
            if lookahead not in "$)":
                lookahead_precedence = self.precedences[lookahead]
                if lookahead_precedence > precedence:
                    rhs = self.parse_binary_operator(rhs, lookahead_precedence)
                else:
                    oper.right = rhs
                    break
            else:
                oper.right = rhs
                break

        return oper
#测试
s = "0-(1+2)*3*(4-3/5/5)+1+1"
print(s)
t = Test(s).parse()
print(t)
print(t.compute(t))

栈调用过深了

以上的解法中生成语法树的过程因为使用了多个递归，结果调用的栈过深，结果 出错了。因为 python默认的递归栈深度是10000。

RecursionError: maximum recursion depth exceeded while calling a Python object

怎么办呢，想办法把递归改成非递归吧。但是看了又看还是没有发现如何改， 因为存在两个函数相互调用的嵌套递归。不管怎样，先减少一个递归函数吧。 去掉了parse_term函数。

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class Test2(object):
    """
    并没有完全把递归改成stack方式。但是因为比Test少了一个parse_term所以减少了递归栈的深度，
    从而侥幸没有栈溢出。
    """

    def __init__(self, s):
        self.s = self.tokenize(s + "$")
        # print(self.s)
        self.i = 0
        self.n = len(self.s)
        self.t = None
        self.precedences = {'+': 0, '-': 0, '*': 1, '/': 1}

    def tokenize(self, s):
        n = len(s)
        result = []
        tmp = ""
        for i in range(0, n):
            char = s[i]
            if char == ' ':
                continue
            if char in '+-*/()$':
                if tmp:
                    result.append(tmp)
                    tmp = ""
                result.append(char)
                continue
            if char in [str(i) for i in range(0, 10)]:
                tmp += char

        return result

    def parse(self):
        return self.parse_expr()

    def parse_expr(self):
        while True:
            char = self.s[self.i]
            t = None
            if char in '$)':
                break
            elif char in '(':
                self.i += 1
                self.t = self.parse_expr()
                c = self.s[self.i]
                if c != ")":
                    raise Exception("{0} shoud be )".format(self.i))
                self.i += 1
            elif char in "+-":
                lhs = self.t
                self.t = self.parse_binary_operator(lhs, self.precedences[char])
            elif char in "*/":
                lhs = self.t
                self.t = self.parse_binary_operator(lhs, self.precedences[char])
            else:
                t = TreeNode(char)
                self.i += 1
                self.t = t

        return self.t

    def parse_binary_operator(self, lhs, precedence):
        c = self.s[self.i]
        oper = TreeNode(c)
        self.i += 1
        oper.left = lhs

        c = self.s[self.i]
        rhs = None
        if c == "(":
            self.i += 1
            t = self.parse_expr()
            # print(t)
            c = self.s[self.i]
            if c != ")":
                raise Exception("{0} shoud be )".format(self.i))
            else:
                self.i += 1
                rhs = t
        else:
            t = TreeNode(c)
            self.i += 1
            rhs = t

        while True:
            lookahead = self.s[self.i]
            if lookahead not in "$)":
                lookahead_precedence = self.precedences[lookahead]
                if lookahead_precedence > precedence:
                    rhs = self.parse_binary_operator(rhs, lookahead_precedence)
                else:
                    oper.right = rhs
                    break
            else:
                oper.right = rhs
                break

        return oper

居然不出现栈溢出了。呵呵。(其实还是存在这个问题，只是能够通过leetcode测试了) 但是计算的过程TreeNode::compute还是出现了调用过深的错误RecursionError。 这个没关系啊，我们可以改成非递归的栈模拟方式。

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class TreeNode(Node):
    def __init__(self, value=None):
        super(TreeNode, self).__init__(value)

    def compute(self, root):
        if root:
            if root.left == None and root.right == None:
                return int(root.value)
            else:

                left = self.compute(root.left)
                right = self.compute(root.right)
                result = None
                char = root.value
                if char == "+":
                    result = left + right
                elif char == "-":
                    result = left - right
                elif char == "*":
                    result = left * right
                elif char == "/":
                    result = left / right

                return result

    def compute2(self, root):
        # compute的非递归形式。
        q = []
        result = None
        prog = 1
        left = None
        right = None
        tup = (prog, root, left, right)
        q.append(tup)
        while q:
            prog, root, left, right = q.pop()
            if prog == 1:
                if root.left == None and root.right == None:
                    result = int(root.value)
                    continue
                else:
                    prog = 2
            if prog == 2:
                tup = (prog + 1, root, left, right)
                q.append(tup)
                tup_new = (1, root.left, left, right)
                q.append(tup_new)
                continue
            if prog == 3:
                left = result
                prog = 4
            if prog == 4:
                tup = (prog + 1, root, left, right)
                q.append(tup)
                tup_new = (1, root.right, left, right)
                q.append(tup_new)
                continue
            if prog == 5:
                right = result
                char = root.value
                if char == "+":
                    result = left + right
                elif char == "-":
                    result = left - right
                elif char == "*":
                    result = left * right
                elif char == "/":
                    result = left / right

        return result

通过的代码

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from binarytree import Node
class TreeNode(Node):
    def __init__(self, value=None):
        super(TreeNode, self).__init__(value)

    def compute(self, root):
        if root:
            if root.left == None and root.right == None:
                return int(root.value)
            else:

                left = self.compute(root.left)
                right = self.compute(root.right)
                result = None
                char = root.value
                if char == "+":
                    result = left + right
                elif char == "-":
                    result = left - right
                elif char == "*":
                    result = left * right
                elif char == "/":
                    result = left / right

                return result

    def compute2(self, root):
        # compute的非递归形式。
        q = []
        result = None
        prog = 1
        left = None
        right = None
        tup = (prog, root, left, right)
        q.append(tup)
        while q:
            prog, root, left, right = q.pop()
            if prog == 1:
                if root.left == None and root.right == None:
                    result = int(root.value)
                    continue
                else:
                    prog = 2
            if prog == 2:
                tup = (prog + 1, root, left, right)
                q.append(tup)
                tup_new = (1, root.left, left, right)
                q.append(tup_new)
                continue
            if prog == 3:
                left = result
                prog = 4
            if prog == 4:
                tup = (prog + 1, root, left, right)
                q.append(tup)
                tup_new = (1, root.right, left, right)
                q.append(tup_new)
                continue
            if prog == 5:
                right = result
                char = root.value
                if char == "+":
                    result = left + right
                elif char == "-":
                    result = left - right
                elif char == "*":
                    result = left * right
                elif char == "/":
                    result = left / right

        return result


class Test2(object):
    """
    并没有完全把递归改成stack方式。但是因为比Test少了一个parse_term所以减少了递归栈的深度，
    从而侥幸没有栈溢出。
    """

    def __init__(self, s):
        self.s = self.tokenize(s + "$")
        # print(self.s)
        self.i = 0
        self.n = len(self.s)
        self.t = None
        self.precedences = {'+': 0, '-': 0, '*': 1, '/': 1}

    def tokenize(self, s):
        n = len(s)
        result = []
        tmp = ""
        for i in range(0, n):
            char = s[i]
            if char == ' ':
                continue
            if char in '+-*/()$':
                if tmp:
                    result.append(tmp)
                    tmp = ""
                result.append(char)
                continue
            if char in [str(i) for i in range(0, 10)]:
                tmp += char

        return result

    def parse(self):
        return self.parse_expr()

    def parse_expr(self):
        while True:
            char = self.s[self.i]
            t = None
            if char in '$)':
                break
            elif char in '(':
                self.i += 1
                self.t = self.parse_expr()
                c = self.s[self.i]
                if c != ")":
                    raise Exception("{0} shoud be )".format(self.i))
                self.i += 1
            elif char in "+-":
                lhs = self.t
                self.t = self.parse_binary_operator(lhs, self.precedences[char])
            elif char in "*/":
                lhs = self.t
                self.t = self.parse_binary_operator(lhs, self.precedences[char])
            else:
                t = TreeNode(char)
                self.i += 1
                self.t = t

        return self.t

    def parse_binary_operator(self, lhs, precedence):
        c = self.s[self.i]
        oper = TreeNode(c)
        self.i += 1
        oper.left = lhs

        c = self.s[self.i]
        rhs = None
        if c == "(":
            self.i += 1
            t = self.parse_expr()
            # print(t)
            c = self.s[self.i]
            if c != ")":
                raise Exception("{0} shoud be )".format(self.i))
            else:
                self.i += 1
                rhs = t
        else:
            t = TreeNode(c)
            self.i += 1
            rhs = t

        while True:
            lookahead = self.s[self.i]
            if lookahead not in "$)":
                lookahead_precedence = self.precedences[lookahead]
                if lookahead_precedence > precedence:
                    rhs = self.parse_binary_operator(rhs, lookahead_precedence)
                else:
                    oper.right = rhs
                    break
            else:
                oper.right = rhs
                break

        return oper

# 测试
s = "0-(1+2)*3*(4-3/5/5)+1+1"
print(s)
t = Test2(s).parse()
print(t)
print(t.compute2(t))

总结

虽然代码长了点，但是练习了一下如何生成一个抽象语法树。并且练习了 如何把一个调用自身的递归改成非递归。

#leetcode #算法 #python
[leetcode 224 & 227]Basic Calculator I & II 原创解法
https://threelambda.com/2016/10/24/leetcode-224/
作者
Ming Yang
发布于
2016年10月24日
许可协议